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Limits: paths of approach

  1. Feb 25, 2014 #1
    1. The problem statement, all variables and given/known data
    By considering different paths, show that the given function has no limit
    as (x,y) [itex]\rightarrow[/itex] (0,0).

    f(x,y) = x4/(x4 + y4)

    2. Relevant equations



    3. The attempt at a solution
    My instructor taught me this process a while back and am unsure if it fits for this problem:

    let y=mx2

    limit as (x, mx2) [itex]\rightarrow[/itex] (0,0) of
    x4/(x4 + mx4)

    I tried this method seeing as letting x=0 yields limit = 0 and letting y = 0 yields limit = 1 ???
     
  2. jcsd
  3. Feb 25, 2014 #2

    Mark44

    Staff: Mentor

    Your substitution is incorrect. If y = mx2, then y4 ≠ mx4.
     
  4. Feb 25, 2014 #3
    Oh right, it would become m2x4

    Therefore it's x4/(x4 + m2x4)

    Factoring out x4 yields 1/(1+m2)

    In similar problems this method proved that the limit depended on the value of m, therefore it did not exist. Is this the case here? I would think so.
     
  5. Feb 25, 2014 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, different values of that limit for different values of "m" mean that the limit depends upon the path. Recall that if a function has a limit then the value of f must be close to that limit. But this shows the value of f close to (0,0) on the line y=2x is different from the value on the line y= 3x.
     
    Last edited: Feb 25, 2014
  6. Feb 25, 2014 #5

    Mark44

    Staff: Mentor

    No, that's wrong as well.
    If y = mx2 (as you have in the OP), then y4 = (mx2)4, right?
     
  7. Feb 25, 2014 #6

    Actually, that was a typo on my part while identifying the problem statement. It is supposed to be
    y2 not y4.

    Thank you for taking the time to catch that mistake though.
     
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